The distance between the line and the point is:
$$ d = \frac{ 2 }{ 5 } $$The distance from the point $ (x_0, y_0) $ to the line $ Ax + By + C = 0 $ is given by:
$$ d = \frac{ \left| Ax_0 + B_yo + C \right| }{ \sqrt{A^2 + B^2}} $$After substituting: $ A = 4 $ , $ B = -3 $ , $ C = -28 $ , $ x_0 = 6 $ and $ y_0 = -2 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 4\cdot6 +\left(-3\right)\cdot\left(-2\right) + \left( -28\right) \right| }{\sqrt{ 4^2 + (-3)^2}} = \\ d =& \frac{ \left| 24 + 6 -28 \right| }{\sqrt{ 16 + 9}} = \\ d =& \frac{ \left| 2 \right| }{\sqrt{ 25}} = \\ d =& \frac{ 2 }{ 5 } = \\ d =& \frac{ 2 }{ 5 } \end{aligned} $$